## 42.26 The key formula

Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Assume $X$ is integral and $\dim _\delta (X) = n$. Let $\mathcal{L}$ and $\mathcal{N}$ be invertible sheaves on $X$. Let $s$ be a nonzero meromorphic section of $\mathcal{L}$ and let $t$ be a nonzero meromorphic section of $\mathcal{N}$. Let $Z_ i \subset X$, $i \in I$ be a locally finite set of irreducible closed subsets of codimension $1$ with the following property: If $Z \not\in \{ Z_ i\}$ with generic point $\xi$, then $s$ is a generator for $\mathcal{L}_\xi$ and $t$ is a generator for $\mathcal{N}_\xi$. Such a set exists by Divisors, Lemma 31.27.2. Then

$\text{div}_\mathcal {L}(s) = \sum \text{ord}_{Z_ i, \mathcal{L}}(s) [Z_ i]$

and similarly

$\text{div}_\mathcal {N}(t) = \sum \text{ord}_{Z_ i, \mathcal{N}}(t) [Z_ i]$

Unwinding the definitions more, we pick for each $i$ generators $s_ i \in \mathcal{L}_{\xi _ i}$ and $t_ i \in \mathcal{N}_{\xi _ i}$ where $\xi _ i$ is the generic point of $Z_ i$. Then we can write

$s = f_ i s_ i \quad \text{and}\quad t = g_ i t_ i$

Set $B_ i = \mathcal{O}_{X, \xi _ i}$. Then by definition

$\text{ord}_{Z_ i, \mathcal{L}}(s) = \text{ord}_{B_ i}(f_ i) \quad \text{and}\quad \text{ord}_{Z_ i, \mathcal{N}}(t) = \text{ord}_{B_ i}(g_ i)$

Since $t_ i$ is a generator of $\mathcal{N}_{\xi _ i}$ we see that its image in the fibre $\mathcal{N}_{\xi _ i} \otimes \kappa (\xi _ i)$ is a nonzero meromorphic section of $\mathcal{N}|_{Z_ i}$. We will denote this image $t_ i|_{Z_ i}$. From our definitions it follows that

$c_1(\mathcal{N}) \cap \text{div}_\mathcal {L}(s) = \sum \text{ord}_{B_ i}(f_ i) (Z_ i \to X)_*\text{div}_{\mathcal{N}|_{Z_ i}}(t_ i|_{Z_ i})$

and similarly

$c_1(\mathcal{L}) \cap \text{div}_\mathcal {N}(t) = \sum \text{ord}_{B_ i}(g_ i) (Z_ i \to X)_*\text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i})$

in $\mathop{\mathrm{CH}}\nolimits _{n - 2}(X)$. We are going to find a rational equivalence between these two cycles. To do this we consider the tame symbol

$\partial _{B_ i}(f_ i, g_ i) \in \kappa (\xi _ i)^*$

see Section 42.5.

Lemma 42.26.1 (Key formula). In the situation above the cycle

$\sum (Z_ i \to X)_*\left( \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{N}|_{Z_ i}}(t_ i|_{Z_ i}) - \text{ord}_{B_ i}(g_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i}) \right)$

is equal to the cycle

$\sum (Z_ i \to X)_*\text{div}(\partial _{B_ i}(f_ i, g_ i))$

Proof. First, let us examine what happens if we replace $s_ i$ by $us_ i$ for some unit $u$ in $B_ i$. Then $f_ i$ gets replaced by $u^{-1} f_ i$. Thus the first part of the first expression of the lemma is unchanged and in the second part we add

$-\text{ord}_{B_ i}(g_ i)\text{div}(u|_{Z_ i})$

(where $u|_{Z_ i}$ is the image of $u$ in the residue field) by Divisors, Lemma 31.27.3 and in the second expression we add

$\text{div}(\partial _{B_ i}(u^{-1}, g_ i))$

by bi-linearity of the tame symbol. These terms agree by property (6) of the tame symbol.

Let $Z \subset X$ be an irreducible closed with $\dim _\delta (Z) = n - 2$. To show that the coefficients of $Z$ of the two cycles of the lemma is the same, we may do a replacement $s_ i \mapsto us_ i$ as in the previous paragraph. In exactly the same way one shows that we may do a replacement $t_ i \mapsto vt_ i$ for some unit $v$ of $B_ i$.

Since we are proving the equality of cycles we may argue one coefficient at a time. Thus we choose an irreducible closed $Z \subset X$ with $\dim _\delta (Z) = n - 2$ and compare coefficients. Let $\xi \in Z$ be the generic point and set $A = \mathcal{O}_{X, \xi }$. This is a Noetherian local domain of dimension $2$. Choose generators $\sigma$ and $\tau$ for $\mathcal{L}_\xi$ and $\mathcal{N}_\xi$. After shrinking $X$, we may and do assume $\sigma$ and $\tau$ define trivializations of the invertible sheaves $\mathcal{L}$ and $\mathcal{N}$ over all of $X$. Because $Z_ i$ is locally finite after shrinking $X$ we may assume $Z \subset Z_ i$ for all $i \in I$ and that $I$ is finite. Then $\xi _ i$ corresponds to a prime $\mathfrak q_ i \subset A$ of height $1$. We may write $s_ i = a_ i \sigma$ and $t_ i = b_ i \tau$ for some $a_ i$ and $b_ i$ units in $A_{\mathfrak q_ i}$. By the remarks above, it suffices to prove the lemma when $a_ i = b_ i = 1$ for all $i$.

Assume $a_ i = b_ i = 1$ for all $i$. Then the first expression of the lemma is zero, because we choose $\sigma$ and $\tau$ to be trivializing sections. Write $s = f\sigma$ and $t = g \tau$ with $f$ and $g$ in the fraction field of $A$. By the previous paragraph we have reduced to the case $f_ i = f$ and $g_ i = g$ for all $i$. Moreover, for a height $1$ prime $\mathfrak q$ of $A$ which is not in $\{ \mathfrak q_ i\}$ we have that both $f$ and $g$ are units in $A_\mathfrak q$ (by our choice of the family $\{ Z_ i\}$ in the discussion preceding the lemma). Thus the coefficient of $Z$ in the second expression of the lemma is

$\sum \nolimits _ i \text{ord}_{A/\mathfrak q_ i}(\partial _{B_ i}(f, g))$

which is zero by the key Lemma 42.6.3. $\square$

Comment #2981 by Xia on

There seems to be some typos in the first displayed formula in the proof of Lemma 41.26.1 and the bracket after it.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).